Question

find the following limit

[(1+2x)^(1/3)-1]/x
lim->0

Answer 1

\[\huge \lim_{x\to 0}\frac{\sqrt[3]{1+2x}-1}{x}\]

I rewrote it better....

you're supposed to know this formula:

\[\Large a^3-b^3=(a-b)(a^2+ab+b^2)\]

...

Answer 2

that's not right kreshnick, you want to normalize the numerator

Answer 3

Thanks for writing it in good form though

Answer 4

@TuringTest what's wrong O_O :(

Answer 5

well I don't see how to do it that way, let's see what you were gonna do

Answer 6

alright... do you have patience to wait till I'll write it with latex?

Answer 7

sure I gotta bounce for a bit anyway, you guys work it out and I'll be happy to see when I get back

Answer 8

alright then ... I might be wrong... but I'll show what I meant

Answer 9

it's even better when I'm wrong, then I get to learn new ways to do things :)

Answer 10

for the record my thought is to multiply by\[{(1+2x)^{2/3}+1\over(1+2x)^{2/3}+1}\]but whatever works :)

Answer 11

sorry to interrupt...is my question right now? (TuringTest)

Answer 12

the way kreshnick has it is the same as yours, if that is what you mean

Answer 13

oh heck, if you can use l'hospital you can do it in a flash

Answer 14

ok brb

Answer 15

without L'hospital please

Answer 16

\[\Large \lim_{x\to 0}\frac{\sqrt[3]{1+2x}-1}{x}\cdot \frac{\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}{\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}\]


\[\Large \lim_{x\to 0}\frac{(\sqrt[3]{1+2x})^3-1^3}{x\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}\]

\[\Large \lim_{x\to 0}\frac{1+2x-1}{x\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}\]

\[\Large \lim_{x\to 0}\frac{2x}{x\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}\]

\[\Large \lim_{x\to 0}\frac{2}{\left[(\sqrt[3]{1+2x}-1)^2+(\sqrt[3]{1+2x}-1)+1^2\right]}\]

\[\Large =\frac{2}{\left[(\sqrt[3]{1+2\cdot 0 }-1)^2+(\sqrt[3]{1+2\cdot 0 }-1)+1^2\right]}=\]

\[\Large =\frac{2}{\left[(1-1)^2+(1-1)+1^2\right]}=\]

\[\Large =\frac{ 2}{1}=2\]

@TuringTest I think I deserve a medal !! LOL

Answer 17

No L'hopital's Rule.. that would be a piece of cake !

Answer 18

thanks man. I knew the formula but, I didn't know how to apply it in this question.

Answer 19

Glad to help ... but hold on a second ! @TuringTest still there?

Answer 20

he is going to come to look at it.

Answer 21

I'm sorry to say that your answer's not quite right :(

Answer 22

try it with l'Hospital and you will see

Answer 23

... ahh , what's wrong :( [ don't be sorry, I'm glad to know that I can learn something ]

Answer 24

not sure, I just got back...
I have to look it over

Answer 25

with L'Hopital's Rule I get 2/3 ... O_O

Answer 26

what do you get ?

Answer 27

the answer is 2/3

Answer 28

I see that you were inconsistent in your use of a in your formula
you made a=(1+2x)^(1/3) at times, and a=(1+2x)^(1/3)-1 at others
I see the logic in your work though, you got the top number right :D

Answer 29

arg, I can't retype the difference in your latex line....

Answer 30

I used .. \[\LARGE a-b= \sqrt[3]{1+2x}\] so I had to multiply by \[\LARGE a^2+ab+b^2\]

to get the third root .....

Answer 31

I GET IT..... I put there -1 too ... pfff... but if that wouldn't be there THIS would be CORRECT ... in the end we get..

\[\LARGE \frac{2}{1+1+1}\]

Answer 32

I though you were saying \(a=\sqrt[3]\)\[\lim_{x\to 0}\frac{\sqrt[3]{1+2x}-1}{x}\cdot \frac{\left[(\sqrt[3]{1+2x})^2+\sqrt[3]{1+2x}+1^2\right]}{\left[(\sqrt[3]{1+2x}^2+(\sqrt[3]{1+2x})+1^2\right]}\]remember that a=

...and I can see you figure out your mistake
nice technique though, new one on me :)

Answer 33

so I just made a typing mistake, this way of doing it works too !

Answer 34

I like being wrong :)

Answer 35

at least I knew what I was doing LOL hahahaha... (ME TOO !)